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    <div class="post-body" itemprop="articleBody"><h1 id="单调栈">单调栈</h1>
<p>栈内元素保持单调递增或单调递减的顺序，常用于解决"寻找最近的比当前元素大/小的元素"这类问题。</p>
<span id="more"></span>
<p>设栈 <span class="math inline">\(S\)</span> 中元素为 <span
class="math inline">\(S_1, S_2, ..., S_k\)</span>，其中 <span
class="math inline">\(S_1\)</span> 为栈底元素，<span
class="math inline">\(S_k\)</span> 为栈顶元素。</p>
<p>以单调递增栈为例，栈中元素满足:</p>
<p><span class="math inline">\(S_1 &lt; S_2 &lt; ... &lt;
S_k\)</span></p>
<p>当需要插入新元素 <span class="math inline">\(x\)</span> 时:</p>
<ul>
<li>当栈非空且 <span class="math inline">\(x &lt; S_k\)</span>
时，不断弹出栈顶元素</li>
<li>直到栈空或 <span class="math inline">\(x \geq S_k\)</span></li>
<li>将 <span class="math inline">\(x\)</span> 压入栈中</li>
</ul>
<p>通过这种方式，栈内元素始终保持单调性。每个元素最多入栈和出栈各一次，因此单调栈的时间复杂度为
<span class="math inline">\(O(n)\)</span>。</p>
<h2 id="例看到牛头">例：看到牛头</h2>
<p>第 <span class="math inline">\(i\)</span> 头牛在队尾，第 <span
class="math inline">\(N\)</span>
头牛在队头，一个牛能向前看到连续的比它矮的牛的牛头，直到有一头牛比它高，再往前的牛头就看不到了。</p>
<p>求每头牛能看到的牛头的数量之和。</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br></pre></td><td class="code"><pre><span class="line">    身高</span><br><span class="line">    ^</span><br><span class="line">    |</span><br><span class="line">15  |                          *</span><br><span class="line">    |                          |</span><br><span class="line">12  |  *                 *     |</span><br><span class="line">    |  |                 |     |</span><br><span class="line">10  |  |  *-------------&gt;|     |</span><br><span class="line">    |  |  |       4      |     |</span><br><span class="line">8   |  |  |        *     |     |</span><br><span class="line">    |  |  |  *     |     |     |</span><br><span class="line">6   |  |  |  |  *  |  *  |     |</span><br><span class="line">    |  |  |  |  |  |  |  |     |</span><br><span class="line">4   |  |  |  |  |  |  |  |  *  |</span><br><span class="line">    |  |  |  |  |  |  |  |  |  |</span><br><span class="line">2   |  |  |  |  |  |  |  |  |  |</span><br><span class="line">    +--+--+--+--+--+--+--+--+--+---&gt; 位置</span><br><span class="line">    1  2  3  4  5  6  7  8  9  10</span><br></pre></td></tr></table></figure>
<p>要快速计算每头牛的“视野”，就要快速知道它前面恰好比它高的最近的那个位置</p>
<p>维护一个单调栈，从<span
class="math inline">\(N\)</span>（队头）开始，序号从大到小逐个处理每头牛身高。</p>
<p>对第 <span class="math inline">\(i\)</span> 头牛身高
<code>h[i]</code>，栈里小于
<code>h[i]</code>的都出栈，直到栈空或遇到一个比 <code>h[i]</code>
高的，这就是挡住第 <span class="math inline">\(i\)</span>
头牛视线的牛，它俩之间就是 <span class="math inline">\(i\)</span>
能看到的牛头个数。而栈里比 <code>h[i]</code> 低的这些，不会影响到 <span
class="math inline">\(j &lt; i\)</span> 那些牛的视野，所以出栈。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdio&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstring&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdlib&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;algorithm&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;stack&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;string&gt;</span></span></span><br><span class="line"><span class="type">const</span> <span class="type">int</span> maxn = <span class="number">8e4</span> + <span class="number">10</span>;</span><br><span class="line"><span class="type">int</span> n, h[maxn];</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;n);</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i ++) &#123;</span><br><span class="line">        <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;h[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    std::stack&lt;<span class="type">int</span>&gt; st;</span><br><span class="line">    <span class="type">long</span> <span class="type">long</span> ans = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = n - <span class="number">1</span>; i &gt;= <span class="number">0</span>; i --) &#123;</span><br><span class="line">        <span class="keyword">while</span>(!st.<span class="built_in">empty</span>() &amp;&amp; h[st.<span class="built_in">top</span>()] &lt; h[i]) &#123;</span><br><span class="line">            st.<span class="built_in">pop</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        ans += st.<span class="built_in">empty</span>() ? n - i - <span class="number">1</span> : st.<span class="built_in">top</span>() - i - <span class="number">1</span>;</span><br><span class="line">        st.<span class="built_in">push</span>(i);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">printf</span>(<span class="string">&quot;%lld\n&quot;</span>, ans);</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="例最大的矩形纸片">例：最大的矩形纸片</h2>
<p><span class="math inline">\(N\)</span>
列，每列给出高度，求最大内部矩形</p>
<img src="/2025-04-05-22-%E5%8D%95%E8%B0%83%E6%A0%88/rec.svg" class="">
<p>分析：每个柱子往一个方向延申，不一定是最优解。最大矩形至少在一根柱子上是抵到顶部的，考虑枚举每个柱子作为抵到顶部的柱子，看往左往右两个方向能延申多远。对于一个方向，单调栈已经可以搞定了，无非就是两个方向分别做一次单调栈。</p>
<p>对于这道题，单调栈维护的是栈底更低、栈顶更高的柱子，对每个位置 <span
class="math inline">\(i\)</span>，栈里比它高的都出栈——因为朝这个方向，比它高的不会影响延申，直到栈空或遇到比它低的，就是延申最远位置了，然后将
<span class="math inline">\(i\)</span> 的高度入栈。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdio&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstring&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdlib&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;algorithm&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;stack&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;string&gt;</span></span></span><br><span class="line"><span class="type">const</span> <span class="type">int</span> maxn = <span class="number">1e6</span> + <span class="number">10</span>;</span><br><span class="line"><span class="type">int</span> n, h[maxn], l_max[maxn];</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;n);</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i ++) &#123;</span><br><span class="line">        <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;h[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    std::stack&lt;<span class="type">int</span>&gt; st_l;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i ++) &#123;</span><br><span class="line">        <span class="keyword">while</span>(!st_l.<span class="built_in">empty</span>() &amp;&amp; h[st_l.<span class="built_in">top</span>()] &gt;= h[i]) &#123;</span><br><span class="line">            st_l.<span class="built_in">pop</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 以 i 为高度往左延申最大矩形长度</span></span><br><span class="line">        l_max[i] = st_l.<span class="built_in">empty</span>() ? i + <span class="number">1</span> : i - st_l.<span class="built_in">top</span>(); </span><br><span class="line">        st_l.<span class="built_in">push</span>(i);</span><br><span class="line">    &#125;</span><br><span class="line">    std::stack&lt;<span class="type">int</span>&gt; st_r;</span><br><span class="line">    <span class="type">long</span> <span class="type">long</span> ans = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = n - <span class="number">1</span>; i &gt;= <span class="number">0</span>; i --) &#123;</span><br><span class="line">        <span class="keyword">while</span>(!st_r.<span class="built_in">empty</span>() &amp;&amp; h[st_r.<span class="built_in">top</span>()] &gt;= h[i]) &#123;</span><br><span class="line">            st_r.<span class="built_in">pop</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 以 i 为高度往右延申最大矩形长度</span></span><br><span class="line">        <span class="type">int</span> r_max = st_r.<span class="built_in">empty</span>() ? n - i : st_r.<span class="built_in">top</span>() - i;</span><br><span class="line">        ans = std::<span class="built_in">max</span>(ans, <span class="number">1LL</span> * (r_max + l_max[i] - <span class="number">1</span>) * h[i]);        </span><br><span class="line">        st_r.<span class="built_in">push</span>(i);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">printf</span>(<span class="string">&quot;%lld\n&quot;</span>, ans);</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

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